Question

Consider thedecay ${\mathit{\Lambda }}^{\mathbf{0}}\mathbf{\to }\mathit{p}\mathbf{+}{\mathit{\pi }}^{\mathbf{-}}$withthe${\mathit{\Lambda }}^{\mathbf{0}}$ at rest. (a) Calculate the disintegration energy. What is the kinetic energy of (b) the proton and (c) the pion?

Short answer

(a) The disintegration energy is.$37.7\text{\hspace{0.17em}Mev}$

(b) The kinetic energy of proton is.$5.35\text{\hspace{0.17em}Mev}$

(c) The kinetic energy of pion is.$32.4\text{\hspace{0.17em}Mev}$

Step-by-step solution

(a) Evaluate disintegration energy.

The given reaction is written as:

${\Lambda }^0\to p+{\pi }^{-}$

Solve the disintegration energy as follows:

$\begin{array}{c}Q=\Delta m{c}^2\\ =(m_{{\Lambda }^0}-m_p-m_{{\pi }^{-}})c^2\end{array}$

Substitute the values from the standard data.

$\begin{array}{c}Q=(1115.6-938.3-139.6)Mev\\ =37.7Mev\end{array}$

Thus, the disintegration energy is.$37.7Mev$

(b) Evaluate the kinetic energy of proton. 

Solve the kinetic energy of proton in the given reaction as follows;

$K_p=\frac{{(E_{{}_{\Lambda }}-E_p)}^2-E_{\pi }^2}{2E_{\Lambda }}$

Substitute the values from the standard data

$\begin{array}{c}{K}_p=\frac{{(1115.6-938.3)}^2-{(139.6)}^2}{2(1115.6)}\text{\hspace{0.17em}Mev}\\ =5.35\text{\hspace{0.17em}Mev}\end{array}$

Thus, the kinetic energy of proton is .$5.35Mev$

(c) Evaluate the kinetic energy of pion.

Solve the kinetic energy of proton in the given reaction as follows;

$\begin{array}{c}{K}_{\pi }=Q-K_p\\ =(37.7-5.35)\text{\hspace{0.17em}Mev}\\ =32.4\text{\hspace{0.17em}Mev}\end{array}$

Thus, the kinetic energy of pion is.$32.4Mev$